SOLUTION: For the function f(x)= (2x^3+3x-5)/(x^2+x-30)
a) The slant asymptote is __________
b) The vertical asymptotes are __________
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Question 389636: For the function f(x)= (2x^3+3x-5)/(x^2+x-30)
a) The slant asymptote is __________
b) The vertical asymptotes are __________
Answer by haileytucki(390) (Show Source): You can put this solution on YOUR website!
f(x)=(2x^(3)+3x-5)/(x^(2)+x-30)
Factor the polynomial using the rational roots theorem.
f(x)=((x-1)(2x^(2)+2x+5))/(x^(2)+x-30)
In this problem 6*-5=-30 and 6-5=1, so insert 6 as the right hand term of one factor and -5 as the right-hand term of the other factor.
f(x)=((x-1)(2x^(2)+2x+5))/((x+6)(x-5))
The domain of an expression is all real numbers except for the regions where the expression is undefined. This can occur where the denominator equals 0, a square root is less than 0, or a logarithm is less than or equal to 0. All of these are undefined and therefore are not part of the domain.
(x+6)(x-5)=0
Solve the equation to find where the original expression is undefined.
x=-6,5
The domain of the rational expression is all real numbers except where the expression is undefined.
x$-6,x$5_(-I,-6) U (-6,5) U (5,I)
The vertical asymptotes are the values of x that are undefined in the function.
x=-6_x=5
A horizontal asymptote can potentially be found by finding the limit of the function as the value approaches I.
L[x:I,((x-1)(2x^(2)+2x+5))/((x+6)(x-5))]
The value of L[x:I,((x-1)(2x^(2)+2x+5))/((x+6)(x-5))] is I.
I
There are no horizontal asymptotes because the limit does not exist.
No horizontal asymptote approaching I.
A horizontal asymptote can potentially be found by finding the limit of the function as the value approaches -I.
L[x:-I,((x-1)(2x^(2)+2x+5))/((x+6)(x-5))]
The value of L[x:-I,((x-1)(2x^(2)+2x+5))/((x+6)(x-5))] is -I.
-I
The horizontal asymptote is the value of y as x approaches -I.
y=-I
Complete the polynomial division of the expression to determine if there is any remainder.
x^(2)+x-30,2x^(3)+0x^(2)+3x-5,-2x^(3)-2x^(2)+60x,M2x^(3)-2x^(2)+63x-5,M2x^(3)-2x^(2)+2x-60,M2x^(3)-2x^(2)-65x-65,2x-2
Split the solution into the polynomial portion and the remainder.
2x-2+(65x-65)/(x^(2)+x-30)
The oblique asymptote is the polynomial portion of the long division result.
y=2x-2
This is the set of all asymptotes for f(x)=((x-1)(2x^(2)+2x+5))/((x+6)(x-5)).
Vertical Asymptote: x=-6,x=5_Horizontal Aysmptote:y=-Infinite_Oblique Aysmptote:y=2x-2
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